Fourth order differential into solving a sum to infinity of a transcendental equation

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I am try to solve this fourth order differential equation numerically:

$EIw''''+w''=-F_0-Pw_0''$

Boundary conditions:

at $x=0, w'=w'''=0$

at $x=\pm L, w=w'=w''=0$

Let:

$w(x)=\sum_{m=1,3,5...}^\infty A_mcos(\frac{m\pi x}{2L})$

$F_0(x)=\sum_{m=1,3,5...}^\infty a_mcos(\frac{m\pi x}{2L})$

$w_0''=\sum_{m=1,3,5...}^\infty b_mcos(\frac{m\pi x}{2L})$

Using Fourier analysis:

$a_m=\frac{4F_0}{m\pi}sin(\frac{m\pi}{2})$

$b_m=-\frac{\pi\delta_0(L/L_0)cos(\frac{m\pi L}{2L})}{(L^2-L^2_0m^2)}$

$A_m=-\frac{(a_m+Pb_m)}{[EI(\frac{m\pi }{2L})^4-P(\frac{m\pi }{2L})^2]}$

It is stated that a complete solution is obtained given that (for x=L):

$w'=-\sum_{m=1,3,5...}^\infty A_m(\frac{m\pi }{2L})sin(\frac{m\pi }{2})=0$

which gives

$w'=-\sum_{m=1,3,5...}^\infty \frac{[\frac{4F_0L^3}{EIm\pi }-\frac{\beta\pi(\delta_0/L_0)cos(\frac{m\pi L_0}{2L})}{(1-\frac{m^2L^2_0}{L^2})}]}{(\frac{m\pi}{2})[(\frac{m\pi}{2})^2-\beta]}sin\frac{m\pi}{2}=0$

where $\beta=\frac{PL^2}{EI}$

L varies between 1 to 20 in increments of 0.25m.

I am unable to solve the equation for $w'=0$, it has been said that $\beta$ is to be determined using a numerical method such as the Newton Raphson method, however I have no idea where to start as it is a transcendental equation and is summing to infinity. In this equation all other variables are known.

Once the equation for $w'=0$ is solved, $P$ is determined and following this $a_m, b_m, A_m$ can be determined.